1) solve:
a) (2n+1C2) = (nC2)sqaured
b) 12(nC2) = n!
If by nCm you mean combinations of n things taken m at a time, you have a problem:
1C2 = 0
Assuming that is what you meant,
2n + 0 = [n(n-1)/2]^2
8n = n^4 - 2n^3 + n^2
n^4 - 2n^3 + n^2 - 8n = 0
Oops. no integer roots.
For the 2nd one,
12(n)(n-1)/2 = n!
6(n^2-n) = n!
Assume n>= 3
n^2 - n - n!/6 = 0
If n = 5, we have
n^2 - n - 20 = 0
(n-5)(n+4) = 0
so, n=5
12(5*4/2) = 120 = 5!
To solve these equations involving combinations and factorials, we need to use the properties of combinatorics and algebra. Let's break down each question step by step:
a) (2n+1C2) = (nC2)^2
To solve this equation, we will use the combination formula:
nCr = n! / (r!(n-r)!)
Here, n and r are non-negative integers, and n! represents the factorial of n.
First, let's solve the left side of the equation:
(2n+1C2) = (2n+1)! / (2!(2n+1-2)!)
= (2n+1)! / (2! * (2n-1)!)
Next, let's solve the right side of the equation:
(nC2)^2 = (n! / (2!(n-2)!))^2
= (n! / (2!(n-2)!))^2
= (n! / (2! * (n-2)!))^2
= (n! / (2! * (n-2)!))^2
= (n! / 2 * (n-2)!)^2
= (n! / (2n-4)!)^2
Since the left and right sides of the equation are equal, we can equate them:
(2n+1)! / (2! * (2n-1)!) = (n! / (2n-4)!)^2
To simplify further, we can cross-multiply:
(2n+1)! * (2n-4)!^2 = (n!)^2 * (2! * (2n-1)!)
We can simplify the factorial terms on both sides of the equation:
(2n+1) * (2n) * (2n-1)! * (2n-4)!^2 = (n!)^2 * 2! * (2n-1)!
Canceling out common terms, we get:
(2n+1) * (2n) * (2n-4)! = (n!) * 2!
Further simplifying:
2n+1 = 2n^2 - 4n + 1
Rearranging the equation, we have:
2n^2 - 4n = 0
Factoring out 2n:
2n(n - 2) = 0
This equation yields two solutions:
1) n = 0
2) n = 2
b) 12(nC2) = n!
Let's use the combination formula to solve this equation:
(nC2) = n! / (2!(n-2)!)
Now, we can rewrite the equation:
12(n! / (2!(n-2)!)) = n!
Simplifying further:
12 * n! / (2!(n-2)!) = n!
After multiplying both sides of the equation by 2!(n-2)!, we have:
12 * n! = n! * (2!(n-2)!)
Canceling out the common factor of n! on both sides:
12 = 2!(n-2)!
Now, we know that 2! = 2, so let's substitute that value:
12 = 2(n-2)!
Divide both sides by 2:
6 = (n-2)!
Since 6 is not a factorial value, there is no integer solution for n in this equation.